AMC 8 · 2001 · #13

Grade 6 arithmeticrate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

fraction-arithmeticratio-proportionmulti-digit-arithmetic identify-subproblems ↑ Prerequisites: fraction-arithmeticratio-proportion

📏 Short solution 💡 2 insights

Problem

Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Richelle's class of $36$ students has $12$ favoring chocolate pie, $8$ apple, and $6$ blueberry. The remaining students split evenly between cherry and lemon. On a pie graph of this data, how many degrees represent the cherry slice?

Givens: Total students: $36$; Chocolate: $12$, apple: $8$, blueberry: $6$; Remaining students split half cherry, half lemon; Answer choices: (A) $10$, (B) $20$, (C) $30$, (D) $50$, (E) $72$

Unknowns: The number of degrees of the cherry slice on the pie graph

Understand

Givens: Total students: $36$; Chocolate: $12$, apple: $8$, blueberry: $6$; Remaining students split half cherry, half lemon; Answer choices: (A) $10$, (B) $20$, (C) $30$, (D) $50$, (E) $72$

Plan

Primary tool: #7 Break Into Subproblems

The question chains two clean steps: first decide how many students prefer cherry pie, then convert that count into a slice of the $360^\circ$ circle. Tool #7 (Break Into Subproblems) keeps the two ideas separate. Subproblem (a) is pure arithmetic — subtract the three known counts from $36$, then halve. Subproblem (b) is one ratio move — take the cherry fraction of $36$ and multiply by $360^\circ$. Keeping them apart avoids the common slip of multiplying by $360^\circ$ before the halving step.

Execute — Answer: D

#7 Break Into Subproblems 4.OA.A.3 Step 1

Subproblem 1a: count the students left after chocolate, apple, and blueberry.
Subtract those three groups from the class total of $36$.

$$36 - 12 - 8 - 6 = 10$$

💡 Take the whole, peel off each known group, and what is left must be the cherry-plus-lemon group.

#7 Break Into Subproblems 4.OA.A.3 Step 2

Subproblem 1b: split the remaining $10$ students in half so cherry and lemon each get the same count.

$$10 \div 2 = 5 \text{ cherry students}$$

💡 "Half prefer cherry and half prefer lemon" is just dividing the leftover by $2$.

#7 Break Into Subproblems 6.RP.A.3 Step 3

Subproblem 2: turn the cherry count into a pie-graph slice.
Cherry covers the fraction $\dfrac{5}{36}$ of the class, so its slice covers the same fraction of the full $360^\circ$ circle.

$$\dfrac{5}{36} \times 360^\circ = 5 \times 10^\circ = 50^\circ \;\Rightarrow\; \textbf{(D)}$$

💡 Because $36 \times 10 = 360$, each student is worth exactly $10^\circ$ on the pie graph, so $5$ cherry students take $50^\circ$.

[1] #7 4.OA.A.3 Subproblem 1a: count the students left after chocolate, apple, and blueberry. Su

[2] #7 4.OA.A.3 Subproblem 1b: split the remaining $10$ students in half so cherry and lemon eac

[3] #7 6.RP.A.3 Subproblem 2: turn the cherry count into a pie-graph slice. Cherry covers the fr

Review

Reasonableness: Check the whole circle adds to $360^\circ$. Each student is worth $10^\circ$, so chocolate $= 120^\circ$, apple $= 80^\circ$, blueberry $= 60^\circ$, cherry $= 50^\circ$, lemon $= 50^\circ$. Sum: $120 + 80 + 60 + 50 + 50 = 360^\circ$. The slices fit the full pie, and cherry equals lemon as required. The trap answers come from skipping the halving (taking $10$ students $\times 10^\circ = 100^\circ$, then halving the degrees instead of the students) or from using $36 \div 360$ instead of $\dfrac{5}{36} \times 360$.

Alternative: Tool #2 (Draw a Picture): sketch the circle and label one student as a $10^\circ$ wedge (since $360 \div 36 = 10$). Then the cherry slice is just $5$ wedges side by side — $5 \times 10^\circ = 50^\circ$. The picture makes the per-student rate obvious and bypasses the fraction step entirely, leading to the same answer (D).

CCSS standards used (min grade 6)

4.OA.A.3 Solve multistep word problems with whole numbers using the four operations (Subtracting the three known pie groups from $36$ and halving the remainder to find that $5$ students prefer cherry.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world problems, including percent and part-of-a-whole (Converting the cherry fraction $\dfrac{5}{36}$ of the class into the slice $\dfrac{5}{36} \times 360^\circ = 50^\circ$ on the pie graph.)

⭐ When the class size divides $360^\circ$ cleanly, each student is worth the same number of degrees on the pie graph. Here $360 \div 36 = 10^\circ$ per student, so the $5$ cherry fans take $50^\circ$ — answer (D).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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